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Question
Factorise: x3−8y3−64z3−24xyz using identity ?
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Solution
Given,x3−8y3−64z3−24xyz=(x)3+(−2y)3+(−4z)3−3(x)(−2y)(−4z)=(x−2y−4z)[x2+(−2y)2+(−4z)2−x(−2y)(−4z)−(−4z)x]=(x−2y−4z)(x2+4y2+16z2+2xy−8yz+4zx).
x3−8y3−64z3−24xyz=(x)3+(−2y)3+(−4z)3−3(x)(−2y)(−4z)=(x−2y−4z)[x2+(−2y)2+(−4z)2−x(−2y)(−4z)−(−4z)x]=(x−2y−4z)(x2+4y2+16z2+2xy−8yz+4zx).
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